CPhill

Consecutive angles of a parallelogram sum to 180°

Let  P  = 5Q

So

5Q + Q  = 180°

6Q = 180

Q = 30°

So P = 5Q  = 5(30)  = 150°

And opposite angles are equal....so......P = R = 150°

Very similar to the last one, RP

Get to the form   (like we did in the other question )

y = mx + b       the slope is  m

A perpendicular line will  have   the slope =   -1/m

Example :

4)   y = -2x  + 4         the slope is  -2 = m

The slope of  a perpendicular line is   -1/m =   -1 / (-2)   =  1/2

So  perpendicular line :   y = (1/2)x

(As a tip....when we have a whole number slope, m....write it as m/1....then "flip it" and make it negative to find the slope of the perpendicular line....so  ..... -2  = -2/1  =   - (1/-2) = 1/2....

When we have a fractional slope....flip it and make it negative to find the slope of the perpendicular line ...so....2/3  = -3/2 )

Can you try the other two, RP???  [ I'll check your answers....if you want ]

OK....good deal....!!!!

We want the form

y = mx + b         .....where m is the slope

So

1. y = 2x + 7         the slope is 2

Equation of a parallel line  (which has the same slope)   (easy)  .... y = 2x - 10

The only difference is the y intercepts....the first one is 7, the second is -10

2.     4y = x - 3         divide both sides by 4

y = (1/4)x - 3/4          the slope  is   (1/4)

Parallel line   :    y = (1/4)x 

3.      2y + 2  = 3x       subtract 2 from both sides

2y = 3x - 2        divide both sides by  2

y = (3/2)x - 1         the slope is 3/2

Parallel line :    y = (3/2)x

Hope that makes sense, RP....!!!!

Momentum = Mass * Velocity

218 kg m / s   =  35. 2 kg * Velocity       divide both sides by  35.2 kg

218 kg m / s

__________ = Velocity  ≈    6.19 m/s

 35.2 kg

I could have taken the first derivative a little farther

-6 ( x^2 + 9)             -6(x^2 + 9)             -6 (x^2 + 9)           -3 (x^2 + 9)

___________ =     ____________ =    ____________ =   _________

(2x^2 - 18)^2         [ 2(x^2 - 9) ]^2         4 (x^2 - 9)^2          2 (x^2 - 9)

This is confirmed by WolframAlpha : https://www.wolframalpha.com/input/?i=derivative++(3x)+%2F+(2x%5E2+-+18)

Regardless.....there are still no max's or mins

I found the first derivative here : https://web2.0calc.com/questions/what-are-the-first-and-second-derivatives-of-the

as

-6 (x^2 + 9) (2x^2 - 18)^(-2)  .........set this to 0

-6 (x^2 + 9)

__________  =     0         multiply each side by (2x^2 - 18)   and we get

(2x^2 -18)^2

-6 (x^2 + 9)  =  0     divide both sides by - 6

x^2 + 9  =   0

Note that the left side is > 0  for all x

Thus....there are no critical points  ( max's or  mins )

Thanks, EP.....Here's another way just using one variable

Let the distance she swims = D

Let the distancvce she runs = 7D

Let the distance she bikes = 3(7D)

So

D + 7D + 3(7D)  = 232    simplify

29D  =  232     divide both sides by 29

D = 8

So....she runs   7D  =  7(8)  =   56 miles

Answered here :  https://web2.0calc.com/questions/what-are-the-first-and-second-derivatives-of-the